Question

How do you find vertical, horizontal and oblique asymptotes for `(2x-2) /( 2x+2 )`?

Answer 1

vertical asymptote x = -1
horizontal asymptote y = 1

Explanation:

Begin by factorising numerator/denominator

`rArr( cancel(2) (x-1))/(cancel(2) (x+1))=(x-1)/(x+1)`

Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation set the denominator equal to zero.

solve: x + 1 = 0 → x = -1 is the asymptote

Horizontal asymptotes occur as `lim_(x to +- oo) , f(x) to 0`

divide terms on numerator/denominator by x

`(x/x-1/x)/(x/x+1/x)=(1-1/x)/(1+1/x)`

as `x to +- oo , f(x) to (1-0)/(1+0)`

`rArry=1" is the asymptote "`

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (both of degree 1) hence there are no oblique asymptotes.
graph{(x-1)/(x+1) [-10, 10, -5, 5]}